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2 years ago

How do you solve this

Mathematics

1 answer:

Trava [24]2 years ago

3 0

1. reduce the fraction and subtract the exponents

(5x^7y^2)/2

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A fish was swimming LaTeX: 2\frac{1}{2}2 1 2 feet below the water's surface at 1:00 p.M. Three hours later, the fish was at a de

marin [14]

Answer:

$\frac{23}{4}$ feet

Step-by-step explanation:

The depth of the fish to the surface of water at 1:00 pm = 2 $\frac{1}{2}$ feet

= $\frac{5}{2}$ feet

The depth of the fish to its initial position at 4:00 pm = 3 $\frac{1}{4}$ feet

= $\frac{13}{4}$ feet

The position of the fish with respect to the water's surface at 4:00 pm = the depth of the fish to the surface of water at 1:00 pm + the depth of the fish to its initial position at 4:00 pm

= $\frac{5}{2}$ + $\frac{13}{4}$

= $\frac{10 + 13}{4}$

= $\frac{23}{4}$

The position of the fish with respect to the water's surface at 4:00 pm is $\frac{23}{4}$ feet.

8 0

2 years ago

There are (43)2 ⋅ 40 strawberries on a farm. What is the total number of strawberries on the farm

WINSTONCH [101]

There are 3,440 strawberries on the farm.

Brainliest please :)

7 0

2 years ago

Read 2 more answers

30 <img src="https://tex.z-dn.net/?f=30%5Cgeq%205" id="TexFormula1" title="30\geq 5" alt="30\geq 5" align="absmiddle" class="lat

Mila [183]

30 is more than 5 not exactlt know what ur looking for here?

3 0

2 years ago

Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and

tankabanditka [31]

So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus $\bf \begin{cases}
b=\textit{rate of the boat}\\
r=\textit{rate of the river}
\end{cases}\qquad thus
\\\\\\

\begin{array}{lccclll}
&distance&rate&time(hrs)\\
&----&----&----\\
upstream&48&b-r&4\\
downstream&48&b+4&3
\end{array}
\\\\\\

\begin{cases}
48=(b-r)(4)\to 48=4b-4r\\\\
\frac{48-4b}{-4}=r\\
--------------\\
48=(b+r)(3)\\
-----------------------------\\\\
thus\\\\
48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)
\end{cases}$

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"

5 0

2 years ago

#1. WILL REWARD BRAINLIEST

Gemiola [76]

c is the answer for 1'

6 0

2 years ago